Question

tan theta =2x(x+1)/2x+1 then find sin theta

Answer 1

Answer:

Sinθ  = 2x(x+1)/(2x² + 2x + 1)

Step-by-step explanation:

Tan theta =2x(x+1)/2x+1 then find sin theta

Tanθ = P/B (Perpendicular /Base)

=> P = 2x(x+1)   & B = 2x+1

H² = P² + B² => H = √ P² + B²

Sinθ = P/H  (Perpendicular /Hypotenuse)

H = √(2x(x + 1))² + (2x + 1)²

= √ 4x²(x² + 1 + 2x)  + (2x + 1)²

=√4x⁴ + 4x²(2x + 1) +  (2x + 1)²

= √(2x²)² + 2*2x²(2x +1)+ (2x + 1)²

= √(2x² + 2x + 1)²

= 2x² + 2x + 1

Sinθ  = 2x(x+1)/(2x² + 2x + 1)

Answer 2

Answer:

tanA = 2x(x+1) / 2x+1 ---- given

sec^2 A = 1+ tan^2 A --- identity.

sec^2 A = 1 + [4x^2(x+1)^2] / (2x+1)^2 = [(2x+1)^2 + 4x^2(x+1)^2] / (2x+1)^2

= [4x^4 + 8x^3 + 8x^2 + 4x +1 ] / (2x+1)^2

= (2x^2 + 2x + 1)^2 / (2x+1)^2 This gives --secA = (2x^2 + 2x + 1) / (2x + 1)

So cosA = (2x + 1) / (2x^2 + 2x + 1)

Now sin^2 A = (1 - cos^2 A ) = 1 - (2x+1)^2 / (2x^2 + 2x + 1)^2

= [(2x^2 + 2x + 1)^2 - (2x+1)^2] / (2x^2 + 2x + 1)^2

= (2x^2)*(2x^2 + 4x + 2 ) / (2x^2 + 2x + 1)^2 ---- using [a^2-b^2 =(a+b)*(a-b) ]

= (4x^2)*(x^2+2x+1) / (2x^2 + 2x + 1)^2 = [(2x )* (x+1)]^2 / (2x^2 + 2x + 1)^2

This gives sinA = (2x )* (x+1) / (2x^2 +2x +1)

These would the corresponding values for sinA and cosA respectively.